Home > Complex questions > Complex #30: sin(z)=z^2 solutions

## Complex #30: sin(z)=z^2 solutions

One of the problems we talked about was the proof of the fact that the equation $\sin(z)=z^2$ has an infinite number of complex number solutions. I thought of a couple of ways to do this. Here is one way.

Let $g(z)=\sin(z)-z^2$. In rectangular coordinates, the function is
$g(x+iy)=\left[\sin(x)\cosh(y)-x^2+y^2\right]+i \left[\cos(x)\sinh(y)-2xy\right]$.

For a large positive integer $N$, consider the rectangle
$R=\left\{ z=x+iy: 2\pi \le x \le 2\pi N, 0\le y \le Y\right\}$, where $Y$ is a large constant to be chosen conveniently later. We will show that as we go counterclockwise around the boundary of $R$, the argument of $g(z)$ increases at least by $2\pi (N-2)$. Then, by the argument principle, the number of zeros inside $R$ is (at least) $(N-2)$ (and there are no poles). So that is all we need for the proof, cause $N$ can be chosen to be arbitrarily large.

Increasing of the argument of $g(z)$:
**On the lower part of the boundary, $y=0$, and
$g(x+i0)=\left[\sin(x)-x^2\right]$. For $x\ge 2\pi$, this is a negative real number, so there is no change in argument as $z$ moves along the lower edge.
**On the left side of the boundary, $x=2\pi$, and
$g(2\pi+iy)=\left[-4\pi^2+y^2\right]+i \left[\sinh(y)-4\pi y\right]$.
As $y$ moves down from (the large) $Y$ to zero, the real part decreases from positive to negative, and the imaginary part also goes from positive to negative to zero. (Note: you can tell that it is still negative near zero by taking a limit  — use Taylor series.) So this tells us that the argument goes from the first to the third quadrant and ends up at the negative real axis, without looping around too much. Without analyzing further, we have no idea if it $g(z)$ goes above or below the origin, so we don’t know if the argument is increasing or decreasing. Either way, we know the argument increases or decreases by at most $\frac{3\pi}{2}$.
**On the right side of the boundary, $x=2N\pi$, and
$g(2N\pi+iy)=\left[-4N^2\pi^2+y^2\right]+i \left[\sinh(y)-4N\pi y\right]$.
As $y$ moves up from  $0$ to (the large) $Y$, the real part goes from negative to positive, and the imaginary part starts at zero, gets negative, then eventually goes positive (since $\sinh(y)$ increases exponentially). So in this case, we move from negative real axis to the first quadrant. Again we can conclude that the argument has changed by no more than $\frac{3\pi}{2}$ (plus or minus).
**On the top part of the boundary, $y=Y$, a large number, and
$g(x+iY)=\left[\sin(x)\cosh(Y)-x^2+Y^2\right]+i \left[\cos(x)\sinh(Y)-2xY\right]$.
As $x$ decreases from $2N\pi$ to $2\pi$, because of the largeness of $Y$, the graph looks like a small perturbation of the graph of $\left[\sin(x)\cosh(Y)\right]+i \left[\cos(x)\sinh(Y)\right]$, which is an ellipse, so we see that the $g(z)$ goes around the origin counterclockwise (notice $\sin$ and $\cos$ switched from usual roles) $(N-1)$ times.

Thus, the total change in argument $\Delta \theta$ of $g(z)$ as $z$ goes counterclockwise around the boundary of $R$ is bounded by $2\pi(N-1)-3\pi\le \Delta\theta \le 2\pi(N-1)+3\pi$, and since
$\Delta\theta$ must be an integral multiple of $2\pi$, we must have that the argument of $g(z)$ makes between $(N-2)$ and $N$ revolutions counterclockwise around the origin. By the argument principle, there must be at least $(N-2)$  zeros inside $R$.

One minor point: how large do we pick $Y$? As long as we choose it so that $\cosh(Y)$ is way bigger than $(2N\pi)^2+Y^2$ and $\sinh(Y)$ is way bigger than $4\pi N Y$, we are fine. And we can definitely pick such a $Y$, since $\sinh(Y)$ and $\cosh(Y)$ increase exponentially in $Y$.

QED